A light rod of length 1.8 m has two point masses of mass 2 kg attached at either

Question

A light rod of length 1.8 m has two point masses of mass 2 kg attached at either end. The rod is pivoted about its centre and is attached to a spring which exerts a moment on the rod that is proportional to the angle the rod is displaced from the equilibrium position and which attempts to return the rod to the equilibrium position. The rod is displaced 0.08 rad from the equilibrium position and released from rest. If the spring constant is 3.7 Nm.rad-1 what is the period of the resulting motion?

[Note: A moment, or more fully a moment of a force, is sometimes called a torque by engineers and those in the US (which is where your textbook was printed).]

Explanation / Answer

Let I is moment of Inertia of the system.

I = 2*m(l/2)^2 = 2*2*0.9^2 = 3.24 Kg m^2

So, w=sqrt(k/I)

=sqrt(3.7/3.24) rad/s

=1.142 rad/s

So, period = 2pi/w = 5.5 s.

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